Let P be a point inside triangle ABC. Rays AP, BP, and CP cut sides BC, CA, and AB at D, E, and F, respectively. Prove that AF/FB + AE/EC = AP/AD.

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This statement may be proved using the area of triangles.

Namely, recall that the area of a triangle is half of the base length multiplied by the corresponding height length. Because of this, when a triangle side (considered as a base) is divided by some ratio, the areas of the...

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This statement may be proved using the area of triangles.

Namely, recall that the area of a triangle is half of the base length multiplied by the corresponding height length. Because of this, when a triangle side (considered as a base) is divided by some ratio, the areas of the corresponding sub-triangles have the same ratio.

In particular, `( AF ) / ( FB ) = | Delta ACF | / | Delta BCF | ` and `( AF ) / ( FB ) = | Delta APF | / | Delta BPF |` where || denotes the area. These two proportions imply that also

`( AF ) / ( FB ) = (| Delta ACF | - | Delta APF |) / (| Delta BCF |-| Delta BPF | ) = | Delta ACP | / | Delta BCP | `

The same way `( AE ) / ( EC ) = | Delta ABP | / | Delta CBP | ` and `( AF ) / ( FB ) + ( AE ) / ( EC ) = (| Delta ACP | + | Delta ABP |) / | Delta BCP |`

Finally, `(AP)/(PD) = |Delta APC|/|Delta DPC| ` and `(AP)/(PD) = |Delta APB|/|Delta DPB|`

Thus,

`( A P ) / ( A D ) = ( | Delta APC | + | Delta APB |) / (|Delta DPC|+|Delta DPB|) = (|Delta APC|+|Delta APB|)/|Delta BCP|`

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